In this section we consider the determinant (REF ) where \(w_k\)  exist and are given by (), with \(I=[a,b]\) , \(0<a<b<1\) . We further assume that \(w\)  does not have Fisher–Hartwig singularities (see [1]}, [2]}, [3]}, or [4]} for instances of Fisher–Hartwig singularities on the real line). Let us again assume that the symbol \(\phi \)  is of Szegő-type. The Riemann–Hilbert approach outlined in this section can be naturally extended to the three other cases: i) \(-1<a<b<0\) , ii) \(-\infty <a<b<-1\) , and iii) \(1<a<b<\infty \) .
We consider the system of orthogonal polynomials \(\lbrace P_n(z)\rbrace \) , \(\deg P_n(z)=n\) , satisfying the following orthogonality conditions
\(\int ^{b}_{a} P_n(x)x^{k+s} w(x)\, {\rm d}x + \int _{ P_n(z)z^{-k-r}\phi (z) \frac{{\rm d}z}{2\pi {\rm i} z} = h_n \delta _{n,k}, \qquad k=0,1,\dots , n.}One can write a determinantal formula for P_n like in (\ref {T+H OP Det rep}), which yields\begin{gather*}h_n=\frac{D_{n+1}(\phi ,w;r,s)}{D_{n}(\phi ,w;r,s)}.\end{gather*}By similar considerations as those mentioned in Section~\ref {Section Hankel on T}, the orthogonal polynomials P_n exist and are unique if D_n \ne 0. Now, assuming that D_n, D_{n-1}\ne 0, we consider the function Y defined as\begin{gather}Y(z;n)\!=\!\begin{pmatrix}P_n(z) & \displaystyle \int ^{b}_{a} \frac{P_n(x)x^sw(x)}{x-z}{\rm d}x + \int _{ \frac{ \tilde{\phi }(\xi )\xi ^{r}\tilde{P}_n(\xi )}{\xi -z} \frac{{\rm d}\xi }{2\pi {\rm i} \xi }\\[2ex]-\displaystyle \frac{P_{n-1}(z)}{h_{n-1}} & -\displaystyle \frac{1}{h_{n-1}} \Bigg \lbrace \int ^{b}_{a} \frac{P_{n-1}(x)x^sw(x)}{x-z}{\rm d}x+ \!\! \int _{ \!\frac{ \tilde{\phi }(\xi )\xi ^r\tilde{P}_{n-1}(\xi )}{\xi -z} \frac{{\rm d}\xi }{2\pi {\rm i} \xi } \Bigg \rbrace }\!,}built from the orthogonal polynomials P_n satisfying (\ref {T+H2 Orthogonality}). Consider the following Riemann--Hilbert problem for finding the 2 \times 2 matrix Y satisfying\begin{itemize}\item \textbf {RH-Y1} \quad Y is holomorphic in ([a,b]).\item \textbf {RH-Y2} \quad For z \in  we have\begin{gather*}Y_+^{(1)}(z;n)=Y^{(1)}_-(z;n),\end{gather*}and\begin{gather*}Y_+^{(2)}(z;n)=Y^{(2)}_-(z;n) + z^{r-1}\tilde{\phi }(z)Y^{(1)}_-\big (z^{-1};n\big ).\end{gather*}\end{itemize}\item \textbf {RH-Y3} \quad For x \in (a,b) we have\begin{gather*}Y_+^{(1)}(x;n)=Y^{(1)}_-(x;n),\end{gather*}and\begin{gather*}Y_+^{(2)}(x;n)=Y^{(2)}_-(x;n) + 2\pi {\rm i} x^s w(x)Y^{(1)}_-(x;n).\end{gather*}\end{pmatrix}\item \textbf {RH-Y4} \quad As z \rightarrow \infty \begin{gather*}Y(z;n)=\bigg ( I + O\bigg (\frac{1}{z}\bigg ) \bigg ) z^{n \sigma _3} = \begin{pmatrix}z^n+\mathcal {O}\big (z^{n-1}\big ) & \mathcal {O}\big (z^{-n-1}\big ) \\\mathcal {O}\big (z^{n-1}\big ) & z^{-n} + \mathcal {O}\big (z^{-n-1}\big )\end{pmatrix}\!,\end{gather*}\end{gather}where Y^{(1)} and Y^{(2)} are the first and second columns of Y, respectively. We have the analogue of Theorem \ref {THM RHP formulation circle} here as well.\) 
